Answer
a) $P_{3}=4392\text{ kPa}$
b) $w_{net,out}=423\text{ kJ/kg}$
c) $\eta_{\text {th }}=56.4\%$
d) $MEP=534\text{ kPa}$
Work Step by Step
(a) Process 1-2: isentropic compression.
$$ \begin{aligned}
T_2 & =T_1\left(\frac{v_1}{v_2}\right)^{k-1}=(300 \mathrm{~K})(8)^{0.4}=689 \mathrm{~K} \\
\frac{P_2 v_2}{T_2} & =\frac{P_1 v_1}{T_1} \longrightarrow P_2=\frac{v_1}{v_2} \frac{T_2}{T_1} P_1=(8)\left(\frac{689 \mathrm{~K}}{300 \mathrm{~K}}\right)(95 \mathrm{kPa})=1745 \mathrm{kPa}
\end{aligned}
$$ Process 2-3: $v=$ constant heat addition. $$
\begin{aligned}
& q_{23 \text { in }}=u_3-u_2=c_v\left(T_3-T_2\right) \\
& 750 \mathrm{~kJ} / \mathrm{kg}=(0.718 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})\left(T_3-689\right) \mathrm{K} \\
& T_3= 1734 \mathrm{~K} \\
& \frac{P_3 v_3}{T_3}=\frac{P_2 v_2}{T_2} \longrightarrow P_3=\frac{T_3}{T_2} P_2=\left(\frac{1734 \mathrm{~K}}{689 \mathrm{~K}}\right)(1745 \mathrm{kPa})=4392\ \mathrm{kPa}
\end{aligned}
$$ (b) Process 3-4: isentropic expansion. $$
T_4=T_3\left(\frac{v_3}{v_4}\right)^{k-1}=(1734 \mathrm{~K})\left(\frac{1}{8}\right)^{0.4}=755 \mathrm{~K}
$$ Process 4-1: $\boldsymbol{v}=$ constant heat rejection. $$
\begin{aligned}
& q_{\text {out }}=u_4-u_1=c_v\left(T_4-T_1\right)=(0.718 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(755-300) \mathrm{K}=327 \mathrm{~kJ} / \mathrm{kg} \\
& w_{\text {het, out }}=q_{\text {in }}-q_{\text {out }}=750-327=423 \mathrm{~kJ} / \mathrm{kg}
\end{aligned}
$$ (c) $\quad \eta_{\text {th }}=\frac{w_{\text {netout }}}{q_{\text {in }}}=\frac{423 \mathrm{~kJ} / \mathrm{kg}}{750 \mathrm{~kJ} / \mathrm{kg}}=56.4 \%$
(d) $$
\begin{aligned}
v_1 & =\frac{R T_1}{P_1}=\frac{\left(0.287 \mathrm{kPa} \cdot \mathrm{m}^3 / \mathrm{kg} \cdot \mathrm{K}\right)(300 \mathrm{~K})}{95 \mathrm{kPa}}=0.906 \mathrm{~m}^3 / \mathrm{kg}=v_{\max } \\
v_{\min } & =v_2=\frac{v_{\max }}{r} \\
\mathrm{MEP} & =\frac{w_{\text {met,out }}}{v_1-v_2}=\frac{w_{\text {netout }}}{v_1(1-1 / r)}=\frac{423 \mathrm{~kJ} / \mathrm{kg}}{\left(0.906 \mathrm{~m}^3 / \mathrm{kg}\right)(1-1 / 8)}\left(\frac{\mathrm{kPa} \cdot \mathrm{m}^3}{\mathrm{~kJ}}\right)=534\ \mathrm{kPa}
\end{aligned}
$$
